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Chapter 2: Problem 4

For \(y^{\prime}+4 y=\frac{4}{3}\) an integrating factor is \(e^{\int 4 d x}=e^{4 x}\) so that \(\frac{d}{d x}\left[e^{4 x} y\right]=\frac{4}{3} e^{4 x}\) and \(y=\frac{1}{3}+c e^{-4 x}\) for \(-\infty < x<\infty .\) The transient term is \(c e^{-4 x}\).

Short Answer

Expert verified
The transient term is \( ce^{-4x} \).

Step by step solution

01

Identify the standard form

The given differential equation is \( y' + 4y = \frac{4}{3} \). Recognize that this is a first-order linear ordinary differential equation of the form \( y' + P(x)y = Q(x) \) where \( P(x) = 4 \) and \( Q(x) = \frac{4}{3} \).
02

Find the integrating factor

Calculate the integrating factor \( \mu(x) = e^{\int P(x) \, dx} = e^{\int 4 \, dx} = e^{4x} \). The integrating factor helps to convert the left side of the differential equation into the derivative of a product.
03

Apply the integrating factor

Multiply the entire differential equation by the integrating factor \( e^{4x} \) to get: \( e^{4x}y' + 4e^{4x}y = \frac{4}{3}e^{4x} \). The left side should now be the derivative of the product \( \frac{d}{dx}[e^{4x}y] \).
04

Solve the derivative equation

Recognize that now \( \frac{d}{dx}[e^{4x}y] = \frac{4}{3}e^{4x} \). Integrate both sides with respect to \( x \). The integral of the left side simplifies to \( e^{4x}y \), and the right side becomes \( \frac{1}{3} e^{4x} + C \) where \( C \) is the constant of integration.
05

Solve for y

To solve for \( y \), divide the entire equation by \( e^{4x} \), obtaining \( y = \frac{1}{3} + Ce^{-4x} \). Here, the general solution includes both particular and complementary solutions.
06

Identify transient term

The transient term in the solution is the part that includes the constant \( C \) which affects the behavior of the solution over time. Thus, it is \( Ce^{-4x} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrating Factor
First-order linear differential equations, like the one given in this exercise, are usually solved through a method involving an integrating factor. This approach simplifies the process by transforming the differential equation into a form that is easier to work with.
  • The integrating factor is a function that, when multiplied with the differential equation, turns the left side into a derivative of a product.
  • In the exercise, the equation is given as \( y' + 4y = \frac{4}{3} \), where the function \( P(x) = 4 \).
  • Therefore, the integrating factor is calculated as \( \mu(x) = e^{\int 4 \, dx} = e^{4x} \).
This calculated integrating factor, \( e^{4x} \), reconfigures the differential equation into a derivative form, specifically \( \frac{d}{dx}[e^{4x}y] \), which is essential for further solving the equation.
General Solution
The general solution of a differential equation represents the complete set of possible solutions, including all particular solutions derived from various initial conditions. For our equation, after applying the integrating factor, we have turned it into a format that allows for straightforward integration.
  • The solution obtained after applying the integrating factor is \( \frac{d}{dx}[e^{4x}y] = \frac{4}{3}e^{4x} \).
  • Integration of both sides leads to \( e^{4x}y = \frac{1}{3} e^{4x} + C \), where \( C \) is a constant of integration.
  • To isolate \( y \), divide through by \( e^{4x} \), yielding \( y = \frac{1}{3} + Ce^{-4x} \).
This expression \( y = \frac{1}{3} + Ce^{-4x} \) forms the general solution, consisting of both the particular solution (\( \frac{1}{3} \)) and the homogeneous solution (\( Ce^{-4x} \)).
Transient Term
The transient term in a differential equation solution highlights transient responses in systems. It represents the portion of the solution that diminishes or grows with time, often decaying to zero, making it temporary or transient.
  • In the provided equation, the transient term is identified as \( Ce^{-4x} \).
  • This term depends on the constant \( C \) and exponentially decays as \( x \) becomes larger.
  • The transient term illustrates how, over time, the solution \( y = \frac{1}{3} + Ce^{-4x} \) stabilizes to \( y = \frac{1}{3} \) as the effect of \( Ce^{-4x} \) diminishes.
Understanding the transient term is crucial as it depicts the transition period or temporary behavior before reaching steady state in many physical and engineering systems.

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Most popular questions from this chapter

(a) Initially the tank contains 300 gallons of solution. since brine is pumped in at a rate of 3 gal/min and the mixture is pumped out at a rate of 2 gal/min, the net change is an increase of 1 gal/min. Thus, in 100 minutes the tank will contain its capacity of 400 gallons. (b) The differential equation describing the amount of salt in the tank is \(A^{\prime}(t)=6-2 A /(300+t)\) with solution $$A(t)=600+2 t-\left(4.95 \times 10^{7}\right)(300+t)^{-2}, \quad 0 \leq t \leq 100$$ as noted in the discussion following Example 5 in the text. Thus, the amount of salt in the tank when it overflows is $$A(100)=800-\left(4.95 \times 10^{7}\right)(400)^{-2}=490.625 \mathrm{lbs}$$. (c) When the tank is overflowing the amount of salt in the tank is governed by the differential equation $$\begin{aligned} \frac{d A}{d t} &=(3 \mathrm{gal} / \mathrm{min})(2 \mathrm{lb} / \mathrm{gal})-\left(\frac{A}{400} \mathrm{lb} / \mathrm{gal}\right)(3 \mathrm{gal} / \mathrm{min}) \\ &=6-\frac{3 A}{400}, \quad A(100)=490.625 \end{aligned}$$ Solving the equation, we obtain \(A(t)=800+c e^{-3 t / 400} .\) The initial condition yields \(c=-654.947,\) so that $$A(t)=800-654.947 e^{-3 t / 400}$$. When \(t=150, A(150)=587.37\) lbs. (d) As \(t \rightarrow \infty,\) the amount of salt is 800 lbs, which is to be expected since \((400 \text { gal })(2 \mathrm{lb} / \mathrm{gal})=800 \mathrm{lbs}\) (e)

Assume that \(A=A_{0} e^{k t}\) and \(k=-0.00012378\). If \(A(t)=0.145 A_{0}\) then \(t \approx 15,600\) years.

For \(0 \leq t \leq 20\) the differential equation is \(20 d i / d t+2 i=120 .\) An integrating factor is \(e^{t / 10},\) so \((d / d t)\left[e^{t / 10} i\right]=\) \(6 e^{t / 10}\) and \(i=60+c_{1} e^{-t / 10} .\) If \(i(0)=0\) then \(c_{1}=-60\) and \(i=60-60 e^{-t / 10} .\) For \(t>20\) the differential equation is \(20 d i / d t+2 i=0\) and \(i=c_{2} e^{-t / 10} .\) At \(t=20\) we want \(c_{2} e^{-2}=60-60 e^{-2}\) so that \(c_{2}=60\left(e^{2}-1\right)\). Thus $$i(t)=\left\\{\begin{array}{ll} 60-60 e^{-t / 10}, & 0 \leq t \leq 20 \\ 60\left(e^{2}-1\right) e^{-t / 10}, & t>20 \end{array}\right.$$

Separating variables and integrating we obtain \\[\frac{d x}{\sqrt{1-x^{2}}}-\frac{d y}{\sqrt{1-y^{2}}}=0 \quad \text { and } \quad \sin ^{-1} x-\sin ^{-1} y=c\\]. Setting \(x=0\) and \(y=\sqrt{3} / 2\) we obtain \(c=-\pi / 3 .\) Thus, an implicit solution of the initial-value problem is \(\sin ^{-1} x-\sin ^{-1} y=\pi / 3 .\) Solving for \(y\) and using an addition formula from trigonometry, we get \\[y=\sin \left(\sin ^{-1} x+\frac{\pi}{3}\right)=x \cos \frac{\pi}{3}+\sqrt{1-x^{2}} \sin \frac{\pi}{3}=\frac{x}{2}+\frac{\sqrt{3} \sqrt{1-x^{2}}}{2}\\].

Solving \(P(5-P)-\frac{25}{4}=0\) for \(P\) we obtain the equilibrium solution \(P=\frac{5}{2} .\) For \(P \neq \frac{5}{2}, d P / d t<0 .\) Thus, if \(P_{0}<\frac{5}{2},\) the population becomes extinct (otherwise there would be another equilibrium solution.) Using separation of variables to solve the initial-value problem, we get \\[ P(t)=\left[4 P_{0}+\left(10 P_{0}-25\right) t\right] /\left[4+\left(4 P_{0}-10\right) t\right] \\] To find when the population becomes extinct for \(P_{0}<\frac{5}{2}\) we solve \(P(t)=0\) for \(t .\) We see that the time of extinction is \(t=4 P_{0} / 5\left(5-2 P_{0}\right).\)
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